1. 题目

Given two integer arrays nums1 and nums2 of length n, count the pairs of indices (i, j) such that i < j and nums1[i] + nums1[j] > nums2[i] + nums2[j].

Return the number of pairs satisfying the condition.

Example 1:
Input: nums1 = [2,1,2,1], nums2 = [1,2,1,2]
Output: 1
Explanation: The pairs satisfying the condition are:
- (0, 2) where 2 + 2 > 1 + 1.Example 2:
Input: nums1 = [1,10,6,2], nums2 = [1,4,1,5]
Output: 5
Explanation: The pairs satisfying the condition are:
- (0, 1) where 1 + 10 > 1 + 4.
- (0, 2) where 1 + 6 > 1 + 1.
- (1, 2) where 10 + 6 > 4 + 1.
- (1, 3) where 10 + 2 > 4 + 5.
- (2, 3) where 6 + 2 > 1 + 5.Constraints:
n == nums1.length == nums2.length
1 <= n <= 10^5
1 <= nums1[i], nums2[i] <= 10^5

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/count-pairs-in-two-arrays
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

2. 解题

• 先进行不等式变换，$nums1[i]-nums2[i]>nums2[j]-nums1[j]$， 两数组做差得到 diff 数组
• $diff[i]+diff[j]>0$，diff 数组排序，对每个 diff[i]，查找 另一个的下限，计算有多少数满足

class Solution {
public:long long countPairs(vector<int>& nums1, vector<int>& nums2) {// 等价于 nums1[i]-nums2[i] > nums2[j]-nums1[j],  i < j// diff[i] + diff[j] > 0, i,j顺序调换也可以满足，所以只需 i!=jint n = nums1.size();vector<int> diff(n);for(int i = 0; i < n; ++i)diff[i] = nums1[i] - nums2[i];sort(diff.begin(), diff.end());long long ans = 0;for(int i = 0; i < n; ++i){auto it = lower_bound(diff.begin(), diff.end(), -diff[i]+1);ans += diff.end()-it;//这么多数满足if(it <= diff.begin()+i)//包含了i，减 1ans--;}return ans/2; // i < j 有一半是重复的}
};

228 ms 92.8 MB C++

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